Is there a finite set of distinct naturals whose reciprocals sum to 1, and no element in the set is one less than a prime?

Question

I'll try to state this formally, forgive me if I botch the notation. Let

$$R=\{n : n+1 \in\mathbb{P}\}.$$ (Where $\mathbb{P}$ is the set of primes)

Then the question is,

$$\exists S \subset (\mathbb{N} \setminus R)\text{ for }s\text{ of finite length, such that }\sum_{s\in S}{\frac{1}{s}}=1 ?$$

It is not hard to find finite subsets of $\mathbb{N}$ whose reciprocals sum to $1$, such as the trivial $\{1\}$, or $\{2,3,6\}$. I suspect that there are infinitely many finite subsets of $R$ with this property, such as $\{2,4,6,12\}$ or $\{2,4,6,18,36\}$.

However, I couldn't find any valid $S$ with a quick search through small numbers, and I'm wondering if it's known whether such a set exists.

Answer 1

We can use the formula $$ \frac1n = \frac1{n+1} + \frac1{n(n+1)} $$ to obtain such a sum. Start with $$ 1 = \frac12 + \frac14 + \frac18 + \frac18 $$ Now, for $n=8$ we can rewrite the last term as $$ \frac18=\frac19+\frac{1}{72} $$ Since $72+1$ is a prime, we have to iterate this, i.e., $$ \frac18=\frac19+\frac{1}{73}+\frac{1}{5256}. $$ Similarly $$ \frac12=\frac13+\frac17+\frac{1}{43}+\frac{1}{1806} $$ and $$ \frac14=\frac15+\frac{1}{20}. $$ Together we obtain $$ 1=\left(\frac13+\frac17+\frac{1}{43}+\frac{1}{1806}\right)+\left(\frac15+\frac{1}{20}\right)+\frac18+\left(\frac19+\frac{1}{73}+\frac{1}{5256}\right). $$ All denominators plus one are not prime numbers.

Answer 2

Your denominators can include any odd number greater than $1$.

Find an odd semiperfect number $n$ (some subset of its proper factors add to $n$). $945$ is the smallest such number. Write $945$ as a sum of (some) of its proper factors. Then divide by $945$.

$945=315+189+135+105+63+45+35+27+15+9+7$.

So $1=\frac{315+189+135+105+63+45+35+27+15+9+7}{945}$.

That is $$1=\frac13+\frac15+\frac17+\frac19+\frac{1}{15}+\frac{1}{21}+\frac{1}{27}+\frac{1}{35}+\frac{1}{63}+\frac{1}{105}+\frac{1}{135}$$

Answer 3

Here's the greedy solution, obtained by repeatedly choosing the largest possible fraction whose denominator isn't one less than a prime.

$$ \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{13} + \frac{1}{93} + \frac{1}{44155} + \frac{1}{8968410360}$$

And note that $8968410361$ is $241 * 347 * 107243$, from Wolfram Alpha.

You can find new terms by editing the following very crude lisp program, which computes the smallest possible denominator for the next term (without checking the primality condition).

(ceiling (/ 1 (- (1- (+ 1/3 1/5 1/7 1/8 1/9)))))